How do you solve Lowest Common Ancestor of Deepest Leaves on LeetCode?

Lowest Common Ancestor of Deepest Leaves (LeetCode #1123) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: Understanding the concept of depth and how it relates to leaves is crucial.

What is the brute force solution for Lowest Common Ancestor of Deepest Leaves?

The brute force approach for Lowest Common Ancestor of Deepest Leaves is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will find all the leaf nodes and their depths first. Then, we will check each node to see if it is the lowest common ancestor of the deepest leaves by checking if it is an ancestor of all deepest leaves. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Lowest Common Ancestor of Deepest Leaves?

Always consider the simplest approach first, even if it's inefficient. Think about how you can reduce the number of passes over the data. Be prepared to explain your thought process clearly and justify your choices.

What are common mistakes when solving Lowest Common Ancestor of Deepest Leaves?

Failing to recognize that the LCA can be found during a single traversal. Not handling edge cases, such as when the tree has only one node.

#1123

Lowest Common Ancestor of Deepest Leaves

Medium

Hash TableTreeDepth-First SearchBreadth-First SearchBinary TreeDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

💡

Intuition

Time O(n)Space O(h)

This approach uses a single DFS traversal to find the deepest leaves and their lowest common ancestor simultaneously. By returning both the depth and the node, we avoid multiple passes over the tree.

⚙️

Algorithm

4 steps

1Step 1: Perform a DFS that returns both the depth of the deepest leaves and the lowest common ancestor of those leaves.
2Step 2: If a node is a leaf, return its depth and itself.
3Step 3: If both left and right children return a depth, the current node is the LCA.
4Step 4: If only one child returns a depth, propagate that child back up.

solution.py23 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def lcaDeepestLeaves(self, root: TreeNode) -> TreeNode:
10        def dfs(node):
11            if not node:
12                return (0, None)
13            left_depth, left_lca = dfs(node.left)
14            right_depth, right_lca = dfs(node.right)
15            if left_depth > right_depth:
16                return (left_depth + 1, left_lca)
17            elif right_depth > left_depth:
18                return (right_depth + 1, right_lca)
19            else:
20                return (left_depth + 1, node)
21
22        return dfs(root)[1]
23

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1Understanding the concept of depth and how it relates to leaves is crucial.
2Recognizing that a single DFS can solve the problem efficiently is key to optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.